Question

There exists a function $f\left(x\right)$ satisfying $f\left(0\right)=1$, $f^{{}^{\prime }}\left(0\right)=-1$, $f\left(x\right)>0$ for all $x$ and

• A. $f^{\prime \prime }\left(0\right)<0$ for all $x$
• B. $-1<f^{\prime \prime }\left(x\right)<0$ for all $x$
• C. $-2\le f^{\prime \prime }\left(x\right)\le -1$ for all $x$
• D. $f^{\prime \prime }\left(x\right)\le -2$ for all $x$

Answer 1

Answer: (A)

$f^{\prime \prime }\left(0\right)<0$ for all $x$

Since $f\left(x\right)$ is continuous and differentiable where $f\left(0\right)=1$
and $f^{\prime }\left(0\right)=-1,f\left(x\right)>0$ for all $x$.
Thus $f\left(x\right)$ is decreasing for $x>0$ and concave downward.
$\Rightarrow f^{\prime \prime }\left(x\right)<0$